does this circuit use conventional current?

[Mike]

In the reader response section in Nuts & Volts last month, a reader
asked if it was possible to detect when someone was on the phone or not
(he had some setup where a computer shared the same line and didn't want
to interrupt the person on the phone).

I'd like to put this circuit together, but not sure why ground is
attached to the output of the NAND gate (rightmost GATE). Is this
circuit using conventional current? Part of me says no because of the
way the LED is connected (current flows from bottom to top), but does it
flow into the output of the NAND gate? :/

I appreciate any help on this. Unfortunately tracing schematics is an
area where I get confused. Are there any good books or Internet sites
that focus on tracing schematics to figure out how they're working?

Here's the picture of the circuit:
http://img208.imageshack.us/img208/7791/busy2wf.jpg

 

[KodKodKod Learning Consulting]

http://www.kodkodkod.com/database.htm

Hi,

This is a Schmitt Trigger NAND gate (4093). The top and bottom
terminals are Vdd and Vss respectively. Left is the input and the
output is connected to the +ve terminal of LED. I don't see that the
current is flowing into the output as LED is a diode. Btw, I think the
current is flowing from top to bottom through the LED.

What do you mean by "conventional current"?

http://www.kodkodkod.com/database.htm

KodKodKod Learning Consulting - Here you get the answers

 

[Mike]

http://www.kodkodkod.com/database.htm

Hi,

This is a Schmitt Trigger NAND gate (4093). The top and bottom
terminals are Vdd and Vss respectively. Left is the input and the
output is connected to the +ve terminal of LED. I don't see that the
current is flowing into the output as LED is a diode. Btw, I think the
current is flowing from top to bottom through the LED.

I thought the LED was forward biased when current flows from ground, to
the cathode, and then into the output (in the case of electron flow).
This is why I thought the circuit was represented using conventional
flow, where electrons flow from positive to negative.

What do you mean by "conventional current"?

I think I botched this term a bit -- sorry. I think it's proper to say
conventional flow.

http://www.kodkodkod.com/database.htm

KodKodKod Learning Consulting - Here you get the answers

Thanks for your help.

 

[John Fields]

http://www.kodkodkod.com/database.htm

Hi,

This is a Schmitt Trigger NAND gate (4093). The top and bottom
terminals are Vdd and Vss respectively. Left is the input and the
output is connected to the +ve terminal of LED. I don't see that the
current is flowing into the output as LED is a diode. Btw, I think the
current is flowing from top to bottom through the LED.

What do you mean by "conventional current"?

http://www.kodkodkod.com/database.htm

KodKodKod Learning Consulting - Here you get the answers

---
If you don't know what's meant by "conventional current", then
what's all this:

"KodKodKod Learning Consulting - Here you get the answers" bullshit
about?

 

[KodKodKod Learning Consulting]

Hi,

Sorry, I didn't read carefully! You used the right term as John Fields
said.

For a diode, the bar is the cathode and the edge of the triangle is the
anode. And current flows from anode to cathode.

LED is just a diode. It can have both hole or electron flows (depends
on how it is made). LED is forward biased when the anode has higher
potential then the cathode and current (conventional) flows from anode
to cathode. For electron, it flows in the opposite direction (from -ve
to +ve as it is attracted by the E-field) and for hoe, it flows from
+ve to -ve as repelled by the E-field. However, since electron has -ve
charge and hole has +ve charge, they contribue to the conventional
current in the same direction.

So when you deal with diode, to determine whether it is forward biased
or reverse biased, it is better to inspect the terminal voltage instead
of how it is connected. In this case, we are sure the anode always has
higher voltage than the cathode (which is connected to the ground). So
the LED is forward biased. (Of course, the anode voltage is changing
during operation and it can be too low (but still > or = ground) to
turn on the LED and it doesn't lid).

Regards,

KodKodKod Learning Consulting

 

[Mike]

Hi,

Sorry, I didn't read carefully! You used the right term as John Fields
said.

For a diode, the bar is the cathode and the edge of the triangle is the
anode. And current flows from anode to cathode.

LED is just a diode. It can have both hole or electron flows (depends
on how it is made). LED is forward biased when the anode has higher
potential then the cathode and current (conventional) flows from anode
to cathode. For electron, it flows in the opposite direction (from -ve
to +ve as it is attracted by the E-field) and for hoe, it flows from
+ve to -ve as repelled by the E-field. However, since electron has -ve
charge and hole has +ve charge, they contribue to the conventional
current in the same direction.

So when you deal with diode, to determine whether it is forward biased
or reverse biased, it is better to inspect the terminal voltage instead
of how it is connected. In this case, we are sure the anode always has
higher voltage than the cathode (which is connected to the ground). So
the LED is forward biased. (Of course, the anode voltage is changing
during operation and it can be too low (but still > or = ground) to
turn on the LED and it doesn't lid).

Regards,

KodKodKod Learning Consulting

Thanks for the explanation of hole/electron flow. So, the answer to my
question is: "Yes, this circuit uses conventional flow"?

 

[Chris]

Thanks for the explanation of hole/electron flow. So, the answer to my
question is: "Yes, this circuit uses conventional flow"?

Hi, Mike. I'm not sure whether you're confident you've gotten a
satisfactory answer yet. Here's a picture that might help (view in
fixed font or M$ Notepad):

|
| ~
| ___ ~ LED1
| .---|___|-->|-----.
| | R1 ~ |
| | ___ ~ LED2 | A |. | K
| o---|___|--|<-----o ------| \|-------
| | R2 | |/ |
| | |
| +| |
| --- |
| - |
| | |
| | |
| '-----------------'
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Looking at the simple circuit diagram, there's a 3V battery and two
resistor-LED strings. In the picture, LED1 will be ON, and LED2 will
be off. If the anode (A) is more positive than the cathode (K), the
diode is said to be forward-biased, like LED1. If the cathode is more
positive than the anode, the diode is said to be reverse-biased. A
forward-biased diode permits current to flow, and a reverse-biased
diode prevents current flow.

I believe you're asking if the output of the last gate is sourcing
current or sinking current. Here are two logic gates, hooked up with
two resistor-LED strings:

|
|
| __ VCC
| -| \ +
| | )o---. |
| -|__/ | V ~
| .-. - ~
| R| | |
| | | .-.
| '-' | |
| | | |
| V ~ __ '-'
| - ~ -| \ |
| | | )o---'
| === -|__/
| GND
|
| Output A Output B
|
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Output A is said to be current-sourcing. Of course a 4093 output can
either source or sink current. When the output is a logic 1 (output
voltage about equal to the battery +), current will flow out of the
gate (thinking conventional current flow), through the LED and
resistor, to COM, and turn on the LED. When the output is logic 0
(output voltage approx equal to the battery -), both sides of the LED
will be at basically the same voltage, so no current will flow.

Output B is said to be current sinking. When the gate output is 1,
both sides of the LED will be at the same voltage, so no current will
flow. But when the gate output becomes 0, current will flow from the
battery + thru the LED and resistor, and into the logic gate output.
The output acts like a sink for current (again, using standard
conventional current flow).

So, in your circuit, the LED will only be on when the NAND gate output
is high. That's a current sinking output.

By the way, don't connect the battery negative or anything else in this
circuit to ground. The symbol is a way of saying circuit common (COM),
or the negative terminal of the battery.

You sound like an intelligent guy, and just need some background in the
basics to be able to have a lot more understanding. I'd recommend
finding a copy of Don Lancaster's "CMOS Cookbook" for more
straightforward, non-technical background on CMOS circuits. Actually,
most of what's in your application circuit is covered in his book.
It's available at many libraries, Amazon, or Mr. Lancaster's website:

http://www.tinaja.com/

Good luck
Chris

 

[Mike]

Hi, Mike. I'm not sure whether you're confident you've gotten a
satisfactory answer yet. Here's a picture that might help (view in
fixed font or M$ Notepad):

|
| ~
| ___ ~ LED1
| .---|___|-->|-----.
| | R1 ~ |
| | ___ ~ LED2 | A |. | K
| o---|___|--|<-----o ------| \|-------
| | R2 | |/ |
| | |
| +| |
| --- |
| - |
| | |
| | |
| '-----------------'
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Looking at the simple circuit diagram, there's a 3V battery and two
resistor-LED strings. In the picture, LED1 will be ON, and LED2 will
be off. If the anode (A) is more positive than the cathode (K), the
diode is said to be forward-biased, like LED1. If the cathode is more
positive than the anode, the diode is said to be reverse-biased. A
forward-biased diode permits current to flow, and a reverse-biased
diode prevents current flow.

I see. That makes sense.

I believe you're asking if the output of the last gate is sourcing
current or sinking current. Here are two logic gates, hooked up with
two resistor-LED strings:

|
|
| __ VCC
| -| \ +
| | )o---. |
| -|__/ | V ~
| .-. - ~
| R| | |
| | | .-.
| '-' | |
| | | |
| V ~ __ '-'
| - ~ -| \ |
| | | )o---'
| === -|__/
| GND
|
| Output A Output B
|
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Output A is said to be current-sourcing. Of course a 4093 output can
either source or sink current. When the output is a logic 1 (output
voltage about equal to the battery +), current will flow out of the
gate (thinking conventional current flow), through the LED and
resistor, to COM, and turn on the LED. When the output is logic 0
(output voltage approx equal to the battery -), both sides of the LED
will be at basically the same voltage, so no current will flow.

Output B is said to be current sinking. When the gate output is 1,
both sides of the LED will be at the same voltage, so no current will
flow. But when the gate output becomes 0, current will flow from the
battery + thru the LED and resistor, and into the logic gate output.
The output acts like a sink for current (again, using standard
conventional current flow).

So, in your circuit, the LED will only be on when the NAND gate output
is high. That's a current sinking output.

That also makes sense. I wasn't aware of those two different terms, so
I'll be sure to look more into that subject. I wasn't aware that current
could flow in to the output of a gate.

By the way, don't connect the battery negative or anything else in this
circuit to ground. The symbol is a way of saying circuit common (COM),
or the negative terminal of the battery.

The ground symbol? Isn't this circuit connecting negative to ground? It
seems like those two sentences are contradictory

You sound like an intelligent guy, and just need some background in the
basics to be able to have a lot more understanding. I'd recommend
finding a copy of Don Lancaster's "CMOS Cookbook" for more
straightforward, non-technical background on CMOS circuits. Actually,
most of what's in your application circuit is covered in his book.
It's available at many libraries, Amazon, or Mr. Lancaster's website:

My background is in software development, but electronics always has
been a part-time hobby that I just never got into deeply. I understand a
lot of the basic principles, including ohm's law, Kirchoff's law,
capacitance, inductance, etc, but one of the things books (at least the
ones I have) don't emphasize well enough is giving examples of breaking
down schematics. Do you have any recommendations for a book devoted to
this subject? I'll check out that CMOS cookbook you mentioned also.
Thanks for taking the time to explain.

 

[Chris]

I see. That makes sense.


That also makes sense. I wasn't aware of those two different terms, so
I'll be sure to look more into that subject. I wasn't aware that current
could flow in to the output of a gate.


The ground symbol? Isn't this circuit connecting negative to ground? It
seems like those two sentences are contradictory


My background is in software development, but electronics always has
been a part-time hobby that I just never got into deeply. I understand a
lot of the basic principles, including ohm's law, Kirchoff's law,
capacitance, inductance, etc, but one of the things books (at least the
ones I have) don't emphasize well enough is giving examples of breaking
down schematics. Do you have any recommendations for a book devoted to
this subject? I'll check out that CMOS cookbook you mentioned also.
Thanks for taking the time to explain.

Hi, Mike. There's always a certain amount of confusion about the word
"ground" with newbies. Many times, "ground" refers to earth ground,
like the green wire on a line cord / "cold water pipe" potential, &c.
This is exactly what you should *not* do here. The only reason why
your circuit will work is that it has a floating 3V battery power
supply. If you connect the "ground" in your circuit to earth ground,
you'll short out the phone TIP/RING as well as probably smoking the
bridge rectifier. Not where you want to go. Just connect the negative
side of the battery to everything in the circuit that shows that
"ground" / circuit common symbol.

The key to reading schematics is finding out the function of all the
elements of the schematic. You can then examine the way they interact.
But knowledge of what the elements do is alwways the first step.

Again, if you want to know about CMOS logic gates (as good a place to
start as any if you know some basics), "CMOS Cookbook" by Lancaster is
just about perfect for a non-technical introduction. As I said, I
believe it covers most all of what you're getting in your circuit. How
to handle funky non-digital inputs, use of schmitt triggers as
oscillators, using gates as buffer/drivers. It could also give you
hints on how to improve your circuit (I'd start by having the output
LED driven by a small logic level MOSFET). And it's a good, lively
read. Among his other talents, Mr. Lancaster was a great columnist for
electronics hobbyist magazines before they all died.

Good luck
Chris

 

[Mike]

Hi, Mike. There's always a certain amount of confusion about the word

"ground" with newbies. Many times, "ground" refers to earth ground,
like the green wire on a line cord / "cold water pipe" potential, &c.
This is exactly what you should *not* do here. The only reason why
your circuit will work is that it has a floating 3V battery power
supply. If you connect the "ground" in your circuit to earth ground,
you'll short out the phone TIP/RING as well as probably smoking the
bridge rectifier. Not where you want to go. Just connect the negative
side of the battery to everything in the circuit that shows that
"ground" / circuit common symbol.

The key to reading schematics is finding out the function of all the
elements of the schematic. You can then examine the way they interact.
But knowledge of what the elements do is alwways the first step.

Again, if you want to know about CMOS logic gates (as good a place to
start as any if you know some basics), "CMOS Cookbook" by Lancaster is
just about perfect for a non-technical introduction. As I said, I
believe it covers most all of what you're getting in your circuit. How
to handle funky non-digital inputs, use of schmitt triggers as
oscillators, using gates as buffer/drivers. It could also give you
hints on how to improve your circuit (I'd start by having the output
LED driven by a small logic level MOSFET). And it's a good, lively
read. Among his other talents, Mr. Lancaster was a great columnist for
electronics hobbyist magazines before they all died.

Good luck
Chris

Thanks again for your help!

 

[Peter Bennett]

Thanks for the explanation of hole/electron flow. So, the answer to my
question is: "Yes, this circuit uses conventional flow"?

No - the answer to the question in the subject line is "any circuit
uses either conventional or electron current, depending on which you
wish to use."

In the Beginning, early scientists studying electricity didn't know
about electrons or other sub-atomic particles, so they somewhat
arbitrarily declared that current was a flow of positive charge from
the positive terminal of the power source, through the external
circuit, and back to the negative terminal of the power source - this
is what we now call "conventional current". Scientists and engineers
have always analysed circuits using conventional current.

In the days of vacuum tubes, electron current was used when teaching
technicians how circuits worked, as you can't readily explain vacuum
tube operation using conventional current.

Circuits are not "designed to use conventional current" or "designed
to use electron current" - the same circuit should result whichever
way you think - as long as you don't change concepts in mid-stream.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Mike]

No - the answer to the question in the subject line is "any circuit
uses either conventional or electron current, depending on which you
wish to use."

In the Beginning, early scientists studying electricity didn't know
about electrons or other sub-atomic particles, so they somewhat
arbitrarily declared that current was a flow of positive charge from
the positive terminal of the power source, through the external
circuit, and back to the negative terminal of the power source - this
is what we now call "conventional current". Scientists and engineers
have always analysed circuits using conventional current.

In the days of vacuum tubes, electron current was used when teaching
technicians how circuits worked, as you can't readily explain vacuum
tube operation using conventional current.

Circuits are not "designed to use conventional current" or "designed
to use electron current" - the same circuit should result whichever
way you think - as long as you don't change concepts in mid-stream.

Thanks for explaining.